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Unformatted text preview: 9—1. A complicated but correct solution to the problem would write all the ﬁelds incident on the
ﬁlm, ﬁnd the intensity, and ﬁnd the ﬁelds transmitted by the hologram. A much simpler
solution is based on Eq. (938) with A2: A1. That equation states that 1 1 1
21; 21, Zr 20 This equation should now be compared with the lens law, which we must adapt to the Sign
convention used in the discussion of holographic image locations. Remembering that zo is
negative for an object to the left of the hologram or lens, the lens law can be written 1 _ 1 + 1
zi f 310 .
Equating these two expressions for 1/ 21-, we see immediately that 1 1 1 1 .._1_ _
“3 f p 27' ZO Zo ' yielding two focal lengths 1 1“1 1 1 2’1
“(and 4‘“ “(‘zfzrzl Note that one of the two lenses has a focal length that depends on the location of the object. 943. (a) We can ﬁnd the maximum spatial frequency with the help of the ﬁgure. The maximum 100m 10011m Object “‘ t / (300)2+(50)2 = 304um 200nm Reference Figure 9-15: . spatial frequency will be the maximum distance from the reference point to any point on
the object (304 11m in this case), divided by A20, 1 - 304nm ‘
m.1.2111.11.W.i.m_.___._______._.__...:1 2 0 1 . ..
fmax A20 1 X10"4M x20 5 00 cyc es/mm on) (b) T he experiment will fail because the periods of all components of the holographic grating
are much smaller than the wavelength of the reconstruction source. As a consequence,
all diffraction orders will be evanescent, and there will be no way to form an image. \’ 9~14.For simplicity, assume that-the refe1ence point is (111 the optical axis The spatial frequency
associated with the fringe pattern generated by interference of this 1eference with 11 pointwsource object distance d eWay ﬁom the reference point will be (1'
fo'm— . ' Equeting f0 to the cutoff frequency f.: (1f each type of ﬁlm and solving for the resulting value
of d we obtain: - . TILX - ' ' I 1 6mm
' _ - I _ . High-Contrast Copy 1 9111111
Ci—Azfcm '30 243 - I _ _ 9.5111111 Agepen FF - ' _ 19.0 mm ...
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- Spring '08